Question

solve the differention equation yzp + zxq= xy ​

Answer 1

CLASSIFICATION OF THE SOLUTION OR INTEGRAL OF PARTIAL DIFFERENTIAL EQUATIONS

A function which satisfies a given partial differential equation is said to be its solution or integral. There are four types of solutions of a partial differential equation.

(a) Complete Integral or Complete Solution

If from the partial differential equation f(x, y, z, p, q) = 0, a relation F(x, y, z, a, b) = 0 is obtained, which contains as many arbitrary constants as there are independent variables in the partial differential equation, then F(x,y, z, a, b) = 0 is called Complete Integral or Complete Solution of the given partial differential equation.

(b) Particular Integral or Particular Solution

If we assign particular values to arbitrary constants a, b in the complete integral, then the solution obtained is called Particular Integral or Particular Solution.

(c) General Integral or General Solution

Let F(x, y, z, a, b) = 0 be the complete solution of a partial differential equation f(x, y, z, p, q)= 0.

If b = (a), then F (x, y, z, a, (a)) = 0 represents one of the families of surfaces given by F(x, y, z, a, b) = 0. The relation between x, y and z obtained by eliminating arbitrary constant 'a' between the equations F(x, y, z, a, 4)(a)) = 0 and = 0 and is called a General Integral or General Solution of the partial differential equation f(x, y, z, p, q) = 0, provided it satisfies the equation.

We have seen Art. 1.4. that if two functions u and v of x, y, z are connected by the relation f(u, v) = 0, then by eliminating f we can obtain a differential equation of the form

Pp + Qq = R.

Obviously, solution of this equation is f(u, v) = 0 which we call the general integral or general solution of the differential equation.

(d) Singular Integral or Singular Solution

Let F (x, y, z, a, b) = 0 be the complete solution of the given partial differential equation f (x, y, z, p, q) = 0. The relation between x, y and z obtained by eliminating the arbitrary constants 'a' and 'b' between the equations

is called Singular Solution or Singular Integral of the partial differential equation f(x, y, z, p, q)=0, provided it satisfies the given equation. The singular solution may or may not be contained in the complete solution of the given partial differential equation.

Note. The relation obtained by eliminating a, b between the equations

is also said to be envelope of the complete integral F(x,y, z, a, b) = 0. Thus, singular solution is the envelope of the complete integral.

SOLUTION OF LINEAR PARTIAL DIFFERENTIAL EQUATIONS BY DIRECT INTEGRATION

Example 1. .

Solution. The given differential equation is

or

Integrating w.r.t. x keeping t constant, we have

[Constant of integration is arbitrary function of t]

Again integrating w.r.t. t, we have

Example 2. given that when .

Solution. Here z is a function of x and y.

Constants will be taken as function of x

Thus the solution is (1)

Now when y = 0, z = ex

From (1), (2)

When

Differentiating (1) w.r.t. y, we have

(3)

Putting y = 0 and in (3), we have

(4)

Solving (2) and (4), we have

From (1), the solution is z = ey cosh x + ey sinh x.

Answer 2

Answer:

The solution of  $yzp + zxq=xy$ differentiation equation  $\phi(c_{1},c_{2}) =0$ will be $\phi (\frac{x}{y} ,xy-z^{2})=0$

Step-by-step explanation:

The given differentiation equation:   $yzp + zxq=xy$

Consider that, $P=xz, Q=yz,R=xy$

Now Langrage's auxiliary equation are:

$\frac{dx}{P} =\frac{dy}{Q} =\frac{dz}{R} \\\frac{dx}{xz} =\frac{dy}{yz} =\frac{dz}{xy}$                            .............(1)

From 1st and 2nd part above equation:

$\frac{dx}{xz} =\frac{dy}{yz} \\\frac{dx}{x} =\frac{dy}{y} \\ \int\limits {\frac{dx}{x}} = \int\limits {\frac{dy}{y}} \\ logx=logy+logc_{1}\\ logx=logyc_{1}\\ x=yc_{1}\\ c_{1}=\frac{x}{y}$                        ..............(2)                  

Choose the multipliers for eq.(1) : y, x, -2z

Then add the nominators and denominators,

$\frac{ydx+xdy-2zdz}{xyz+xyz-2xyz}$

$\int\limit {{ydx+xdy-2zdz}}=0\\ \int (dx y-2zdz)=0\\ \int{dxy}-2\int{zdz}=0\\ xy-2\frac{z^{2}}{2}=c_{2}\\ xy=z^{2}=c_{2}$                   .............(3)

The general solution is   $\phi (c_{1},c_{2})=0$

Therefore, solution of given differential equation is  $\phi (\frac{x}{y}, xy-z^{2})=0$.